Let's say I have an operator $Y$ which can be expanded in powers of small $\varepsilon$ such that $Y$ to lowest order is already of order $\varepsilon$. I now want to find the expression for $\log(e^X e^Y e^X)$ to linear order in $Y$. Is there a closed expression for this? It's basically the symmetric BCH-formula but it seems like there will always be nested commutators with one $Y$ at arbitrary order, is this true?
I first came up with the following, by making use of \begin{equation} \log(e^X e^Y) = X + \frac{\text{Ad}_X}{1 - \exp(-\text{Ad}_X)}Y + \mathcal{O}(Y^2) \end{equation} Now I want to apply this to another $X$. To have the expression with the expansion parameter hidden in $Y$ on the right again, we first use that $\log(e^X e^Y) = - \log(e^{-X} e^{-Y})$. Eventually, I got something of the form \begin{equation} \log(e^X e^Y e^X) = 2X - \frac{\text{Ad}_X}{1 - \exp(\text{Ad}_X)} \frac{\text{Ad}_X}{1 - \exp(-\text{Ad}_X)}(Y + \mathcal{O}(Y^2)) + \mathcal{O}\left(\left(X + \frac{\text{Ad}_X}{1 - \exp(-\text{Ad}_X)}Y\right)^2\right) \label{eq} \end{equation} The problem is that there might be terms containing only one $Y$ in the last term, which we don't know.
I applied the following reason to say that we can ignore those terms, but I don't know if it's true. In general, we can write those terms as something of the form \begin{equation} f(\text{Ad}_X)(X + Y')^n \end{equation} with $f(x)$ some function whose expansion does not start at $1$, and where $Y'$ is just shorthand notation and contains at least one $Y$, so we really want to expand in $Y'$ here. So, to first order we can just look at \begin{equation} f(\text{Ad}_X)(X^n + P(X...XY') + \mathcal{O}(Y'^2)) \end{equation} With $P(X...XY')$ being the sum over all permutations of one $Y'$ with $n-1$ times $X$. Obviously, because the adjoint action is with $X$, this reduces to \begin{equation} P(X...X (f(\text{Ad}_X)Y')) \end{equation} This means that we have a sum over a bunch of terms where $n-1$ $X$-operators do not appear within nested commutators. However, we know that the only terms not appearing in a nested commutator should be $X + Y$, which is essentially captured in the first term of the expansion of the second term in $\log(e^X e^Y e^X)$. So, we can ignore these terms as they should vanish.
Is this a correct assumption? I have the feeling they might combine with higher-order terms to give nested commutators again, but I am not sure. Any help with these steps, or resources in other directions for my question is highly appreciated.
(Edit:) Doing the computation by hand up to $\text{Ad}^4_X(Y)$ leads to the expression \begin{equation} \log(e^X e^Y e^X) \approx 2X + Y - \frac{1}{12} \text{Ad}^2_X(Y) + \frac{1}{240} \text{Ad}^4_X(Y) \end{equation} These are indeed the correct coefficients if you expand my proposed formula with $\frac{-x^2}{(1-e^{-x})(1-e^x)}$ up to the same order.